Solving Chemistry Problem II: 0.005 M Carbonate Solution

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Discussion Overview

The discussion revolves around a chemistry problem involving the calculation of the mass of an unknown sample required to prepare a 15 mL solution containing 0.005 M carbonate, given that the weight percent of CaCO3 in the sample is 50%. The scope includes mathematical reasoning and problem-solving related to molarity and mass calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant asks for help in calculating the mass of an unknown sample needed to prepare a specific molarity of carbonate solution, noting that half of the sample consists of CaCO3.
  • Another participant suggests determining the number of moles of CaCO3 required for the desired concentration in the solution.
  • A subsequent post calculates the moles of carbonate needed, resulting in 7.5 E-5 moles of CaCO3.
  • Another participant prompts further action by asking how to convert moles to mass using the molar mass equation.
  • There is a mention that the mass of CaCO3 represents 50% of the total mass of the unknown sample.

Areas of Agreement / Disagreement

Participants are engaged in a collaborative problem-solving process, with no explicit consensus reached on the final steps or calculations needed to arrive at the solution.

Contextual Notes

Participants have not fully resolved the mathematical steps required to find the total mass of the unknown sample, nor have they clarified the molar mass of CaCO3 needed for the calculations.

higherme
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My question is:

Assuming the weight percent of CaCO3 is 50%, calculate the mass of unknown sample required to prepare a 15 mL solution that contains 0.005 M carbonate.

Well i know that in the sample of unknown, there half of them are CaCO3. I don't know what do to next...help please

Thank you
 
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if you mean the mass percent of CaCO3 is 50% in an unknown sample then, try getting to how many moles of CaCO3 you would need to make a .005 M solution of Carbonate in 15mL
 
that means i would need:
0.015 L x 0.005 mol/L = 7.5 E -5 moles of carbonate which means i neeed 7.5 E-5 moles of CaCO3

then... ?
 
keep going, how do you get mass (grams) from moles? a great equation to know is MM (molar mass) = g/mol

then you have mass of CaCO3, if its 50% (or 1/2) of the mass (x) your unknown sample ...
 

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